TRIGONOMETRY

TRIGONOMETRY. Nature and Construction of Trigonometrical Tables. TRIGONOMETRY is the application of arithmetic to geometry. It consists of two principal parts, viz. PLANE TRIGONOMETRY and SPHERICAL TRIGONOMETRY. Plane trigonometry treats of the application of numbers to determine the relations of the sides and angles of a plane triangle to one another. Spherical trigonometry treats of the application of numbers in like manner to spherical triangles; the nature of these will be explained in the course of this article. Both branches of the subject depend essentially upon certain numerical tables, the nature and construction of which we shall now proceed to explain. SECTION I. NATURE AND CONSTRUCTION OF TRIGONOMETRICAL TABLES. It has been demonstrated in GEOMETRY (Theor. 31. Sect. IV.) that any angles at the centre of a circle have to one another the same proportion as the arcs intercepted between the lines which contain the angles. Hence it is easy to infer, that an angle at the centre of a circle has the same ratio to four right angles, that the arc intercepted between the lines which contain the angle has to the whole circumference. It also follows that we may employ arcs of a circle as measures of angles, and thus the comparison of angles is reduced to the comparison of arcs of a circle. From this principle we infer the consistency of the first of the following series of definitions. DEFINITIONS. I. If two straight lines intersect one another in the centre of a circle, the arc of the circumference intercepted between them is called the Measure of the angle which they contain. Thus, (Plate DXXXVII. fig. 1.) the arc AB is the measure of the angle contained by the lines CA and CH. II. If the circumference of a circle be divided into 360 equal parts, each of these is called a Degree; and if a degree be divided into 60 equal parts, each of these is called a Minute; and if a minute be divided into 60 equal parts, each of these is called a Second, and so on; and as many degrees, minutes, seconds, &c. as are in any arc, so many degrees, minutes, seconds, &c. are said to be in the angle measured by that arc. COR. 1. Any arc is to the whole circumference of which it is a part, as the number of degrees and parts of a degree in it is to the number 360. And any angle is to four right angles as the number of degrees, &c. in the arc which is the measure of the angle to 360. COR. 2. Hence also it appears that the arcs which measure the same angle, whatever be the radii with which they are described, contain the same number of degrees and parts of a degree. The degrees, minutes, seconds, &c. contained in an arc or angle are commonly written thus, 23^{\circ} 29' 32'', which expression means 23 degrees 29 minutes 32 seconds and 20 thirds. III. Two angles which make together two right angles, also two arcs which make together a semi-circle, are called the Supplements of one another. IV. A straight line BG drawn through B, one of the extremities of the arc AB, perpendicular to the diameter passing through the other extremity A, is called the Sine of the arc AC, or of the angle ACB, having arc AB for its measure. COR. 1. The sine of a quadrant or of a right angle is equal to the radius. COR. 2. The sine of an arc is half the chord of twice the arc. V. The segment AG of the diameter intercepted between its extremity and the sine BG is called the Verfed Sine of the arc AB, or of the angle ACB. VI. A straight line AH touching the circle at A one extremity of the arc AB, and meeting the diameter CB which passes through B the other extremity, is called the Tangent of the arc AB, or of the angle ACB. COR. The tangent of half a right angle is equal to the radius. VII. The straight line CH between the centre and the extremity of the tangent AH is called the Secant of the arc AB or of the angle ACB. COR. to Def. 4, 6, 7. The sine, tangent, and secant of any angle ACB, are also the sine, tangent, and secant of its supplement BCE. For by the definition, BG is the sine of the angle BCE; and if BC be produced to meet the circle in I, then AH is the tangent and CH the secant of the angle ACB or BCE. COR. to Def. 4, 5, 6, 7. The sine, versed sine, tangent, and secant of an arc which is the measure of the angle ACB is to the sine, versed sine, and secant of any other arc which is the measure of the same angle as the radius of the first arc is to the radius of the second. Let BG, fig. 2. be the sine, AG the versed sine, AH the tangent, and CH the secant of the arc AB to the radius CA; and bg, ag, ah, ch the same things to the radius Co. From similar triangles BG: bg:: BC: bC; and because CG: Cg:: CB: Cb:: CA: Ca; therefore, by division AG: ag:: CA: Ca. Also AH: ah:: CH: Ch:: CA: Ca. Hence it appears that if tables be constructed exhibiting in numbers the sines, tangents, and versed sines of certain angles to a given radius, they will exhibit the ratios of the sines, tangents, and versed sines of the same angles to any radius whatever. In such tables, which are called trigonometrical tables, the radius is either supposed 1, or some number in the series 10, 100, 1000, &c. Nature and Construction of Trigonometrical Tables. The construction and use of these tables we shall presently explain. VIII. The difference between any angle and a right angle, or between any arch and a quadrant, is called the Complement of that angle, or of that arch. Thus, if the angle ACD, fig. 1. be a right angle, and consequently the arch AD, which is its measure, a quadrant, the angle BCD is the complement of the angle BCA, and the arch BD is the complement of the arch AB. Also the complement of the obtuse angle BCE is BCD, its excess above a right angle; and the complement of the arch BDE is the arch BD. IX. The sine, tangent, or secant of the complement of any angle is called the cosine, cotangent, or cosecant of that angle. Thus, supposing the angle ACD to be a right angle, then BF=CG, the sine of the angle BCD, is the cosine of the angle BCA; DK, the tangent of the angle BCD, is the cotangent of the angle BCA, and CK, the secant of the angle BCD, is the cosecant of the angle BCA. The following properties of the lines which have been defined flow immediately from their position. 1. The sum of the squares of the sine and cosine of any angle is equal to the square of the radius. For, in the right-angled triangle BGC, BC^2 = BG^2 + GC^2, (GEOMETRY, Sect. IV. theor. 13.) Now BG is the sine, and CG=BF is the cosine of the angle BCA. 2. The radius is a mean proportional between the tangent of any angle and its cotangent, or \tan. ACB \times \cot. ACB = \text{rad.}^2. For since DK, CA are parallel, the angles DKC, HCA are equal; now CDK, CAH are right angles, therefore the triangles CDK, CAH are similar, and therefore AH: AC:: CD or AC: DK, and AC^2 = AH \times DK. 3. The radius is a mean proportional between the cosine and secant of any angle. Or \cos. ACB \times \sec. ACB = \text{rad.}^2. For the triangles CGB, CAH are similar; therefore CG: CB or CA:: CA: CH. 4. The tangent of an arch is a fourth proportional to its cosine, its sine and the radius, or \tan. ACB = \frac{\sin. ACB}{\cos. ACB} \times \text{rad.}. For, from similar triangles CG: GB:: CA: AH. Trigonometrical tables usually exhibit the sines, tangents, and secants of all angles which can be expressed by an exact number of degrees and minutes from 1 minute to 90 degrees, or a right angle. These may be computed in various ways, the most elementary is to calculate them by the help of principles deducible immediately from the elements of geometry. It has been demonstrated in GEOMETRY, (Sect. V. prob. 22.) that the chord of one-sixth of the circumference, or an arch of 60^\circ, is equal to the radius; therefore, if BD be an arch of 30^\circ, its sine BF will be half the radius (cor. 2. def. 4.). Let us suppose the radius to be expressed by unity, or 1, then \sin. 30^\circ = \frac{1}{2}; now since a being put for any arch, \cos. a + \sin. a = \text{rad.}^2 (where by \cos. a is meant the square of the number expressing the cosine of the arch a, &c.) and as \sin. a = \frac{1}{2}, therefore \cos. a = 1 - \frac{1}{2} = \frac{1}{2}, &c. \cos. 30^\circ = \frac{1}{2}\sqrt{3} =.8660254038. It has been demonstrated in the arithmetic of sines (ALGEBRA, § 356.) that 2 \cos. a = 1 + \cos. 2a; hence we have the following formula for finding the cosine of an arch, having given the cosine of its double; \cos. a = \frac{1 + \cos. 2a}{2}. By this formula from the cosine of 30^\circ we may find that of 15^\circ, and again from \cos. 15^\circ we may find \cos. 7^\circ 30', and proceeding in this way we may find the cosines of 3^\circ 45', 1^\circ 52' 30'', and so on, till after 11 bisections the cosine of 52'' 44''' 3^{11} 45'' is found; we may then find the sine of this arch by the formula \sin. a = \sqrt{1 - \cos. a}. Now, as from the nature of a circle the ratio of an arch to its sine approaches continually to that of equality, when the arch is continually diminished, it follows that the sines of very small arches will be very nearly to one another as the arches themselves: Therefore, as 52'' 44''' 3^{11} 45'' to 1' so is the sine of the former arch to the sine of the latter. By performing all the calculations which we have here indicated, it will be found that the sine of 1' is.0002908882. It has been shewn in the arithmetic of sines (ALGEBRA, § 355.) that a and b being put for any two arches, \sin. (a+b) = 2 \cos. b \sin. a - \sin. (a-b), hence putting 1' for b, and 1', 2', 3', &c. successively for a, we have, \begin{aligned} \sin. 2' &= 2 \cos. 1' \times \sin. 1', \\ \sin. 3' &= 2 \cos. 1' \times \sin. 2' - \sin. 1', \\ \sin. 4' &= 2 \cos. 1' \times \sin. 3' - \sin. 2', \\ &\&c. \end{aligned} In this way the sines for every minute of the quadrant may be computed, and as the multiplier \cos. 1' remains always the same, the calculation is easy. If instead of 1', the common difference of the series of arches were any other angle, the very same formula would apply. The sines, and consequently the cosines of any number of arches being supposed found, their tangents may be found by considering that \tan. a = \frac{\sin. a}{\cos. a}; and their secants from the formula \sec. a = \frac{1}{\cos. a}. We have here very briefly indicated the manner of constructing the trigonometrical canon, as it is sometimes called. There are, however, various properties of sines, tangents, &c. which greatly facilitate the actual calculation of the numbers, these the reader will find detailed in ALGEBRA, Sect. XXV. which treats expressly of the Arithmetic of Sines. The most expeditious mode of computing the sine or cosine of a single angle is by means of infinite series: The investigation of these is given in FLUXIONS, § 70. and it is there shewn that if a denote any arch, then, the radius being expressed by 1, \sin. a = a - \frac{a^3}{1 \cdot 2 \cdot 3} + \frac{a^5}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5} - \&c. \cos. a = 1 - \frac{a^2}{1 \cdot 2} + \frac{a^4}{1 \cdot 2 \cdot 3 \cdot 4} - \&c. To apply these we must have the arch expressed in parts of the radius, which requires that we know the proportion of the diameter of the circle to its circumference. We have investigated this proportion in GEOMETRY, Prop. 6. Sect. vi.; also in FLUXIONS, § 137. and subsequently in the article entitled SQUARING THE CIRCLE. From these series others may be found which shall express the tangent and secant. Thus because \tan. a = Nature and Construction of Trigonometrical Tables. a = \frac{\sin. a}{\cos. a}, we get, after dividing the series for the fine by that for the cosine, \tan. a = a + \frac{a^3}{3} + \frac{2a^5}{15} + \frac{17a^7}{315} + \&c. And in like manner, dividing unity by the series for \cos. a, because \sec. a = \frac{1}{\cos. a}, we get \sec. a = 1 + \frac{a^2}{2} + \frac{5a^4}{24} + \frac{61a^6}{720} + \&c. We shall conclude what we proposed to say on the construction of the tables, by referring such of our readers as wish for more extensive information on this subject to Dr Hutton's Introduction to his excellent Mathematical tables; also to the treatises which treat expressly of trigonometry, among which are those of Emerson, Simpson, Bonnycastle, Cognoli, Mauduit, Lacroix, Legendre. In particular, we refer to an excellent treatise on the subject by Mr R. Woodhouse of Caius college, Cambridge. Description of the Table of Logarithmic Sines, &c. That trigonometrical tables may be extensively useful, they ought to contain not only the fine, tangent, and secant to every minute of the quadrant, but also the logarithms of these numbers; and these are given in Dr Hutton's Mathematical Tables, a work which we have already mentioned; as, however, the fines, &c. or the natural fines, &c. as they are called, are much less frequently wanted than their logarithms, we have only given a table of the latter. See LOGARITHMS. This table contains the logarithms of the fines and tangents, or the logarithmic fines and tangents, to every minute of the quadrant, the degrees at top and minutes descending down the left-hand side, as far as 45^\circ, and from thence returning with the degrees at the bottom and the minutes ascending by the right hand side to 90^\circ, in such a manner that any arch on the one side is in the same line with its complement on the other, the respective fines, cosines, tangents, and cotangents, being in the same line with the minutes, and on the columns figured with their respective names at top when the degrees are at top, but at the bottom when the degrees are at the bottom. The differences of the fines and cosines are placed in columns to the right-hand, marked D; and the differences of the tangents and cotangents are placed in a column between them, each difference belonging equally to the columns on both sides of it. Also each differential number is set opposite the space between the numbers whose difference it is. All this will be evident by inspecting the table itself. There are no logarithmic secants in the table, but these are easily had from the cosines; for since \sec. a = \frac{\text{rad.}}{\cos. a}, therefore, \log. \sec. a = 2 \log. \text{rad.} - \log. \cos. a; now \log. \text{rad.} = 10, therefore the log. secant of any arch is had by subtracting its log. cosine from 20. The log. fine, log. tangent, or log. secant of any angle is expressed by the same numbers as the log. fine, log. tangent, or log. secant of its supplement; therefore, when an angle exceeds 90^\circ, subtract it from 180^\circ and take the log. fine, &c. of the remainder for that of Nature and Construction of Trigonometrical Tables. To find the log. fine of any angle expressed by degrees and minutes. If the angle be less than 45^\circ, look for the number of degrees at the top, and opposite to the minutes on the left hand will be found the fine required; thus the log. fine of 8^\circ 10' is 9.15245. But if the angle be 45^\circ or more than 45^\circ, look for the degrees at the bottom and the minutes on the right hand, and opposite will be found the log. fine required. Thus the log. fine of 58^\circ 12' is 9.92936. The very same directions apply for the cosine, tangent, and cotangent; and from what has been said, the manner of finding the angle to degrees and minutes, having given its fine, &c. must be obvious. If the angle consists of degrees, minutes, and seconds, find the fine or tangent to the degrees and minutes, and add to this a proportional part of the difference given in the column of differences for the seconds, observing that the whole difference corresponds to 1' or 60''. Thus to find the log. fine of 30^\circ 23' 28''; first the fine of 30^\circ 23' is 9.70396. The difference is 21. As 60'': 28'':: 21: \frac{28 \times 21}{60} = 10 nearly, the part of the difference to be added, therefore the fine of 30^\circ 23' 28'' is 9.70406. On the contrary, let it be required to find the angle corresponding to the tangent 10.14152. The next less tangent in the table is 10.14140, which corresponds to 54^\circ 10'; the difference between the proposed tangent and next less is 12; and the difference between the next less and next greater, as given in the table, is 26; therefore, 26: 12:: 60'': \frac{12 \times 60}{26} = 28'' nearly, hence the angle corresponding to the proposed log. tangent is 54^\circ 10' 28''. SECTION II. PLANE TRIGONOMETRY. THE following propositions express as many of the properties of plane triangles as are essentially necessary in plane trigonometry. THEOR. I. In a right-angled plane triangle, as the hypothenuse is to either of the sides, so is the radius to the fine of the angle opposite to that side; and as either of the sides to the other side, so is the radius to the tangent of the angle opposite to that side. Let ABC be a right-angled plane triangle (fig. 3.), of which AC is the hypothenuse. On A as a centre with any radius, describe the arch DE; draw EG at right angles to AB, and draw DF touching the circle at D, and meeting AC in F. Then EG is the fine of the angle A to the radius AD or AE, and DF is its tangent. The triangles AGE, ADF are manifestly similar to the triangle ABC. Therefore AC: CB:: AE: EG; that is, AC: CB:: \text{rad.}: \sin. A. Again, Again, AB: BC:: AD: DF; that is AB: BC::rad.: tan. A. COR. In a right-angled triangle, as the hypotenuseto either of the sides, so is the secant of the acute angleadjacent to that side to the radius. For AF is the secantof the angle A to the radius AD; and AC: AB:: AF: AD, that is, AC: AB:: \text{sec. A}: \text{rad.} Note. This proposition is most easily rememberedwhen stated thus. If in a right-angled triangle the hypo-thenuse be made the radius, the sides become the fines ofthe opposite angles; and if one of the sides be made theradius, the other side becomes the tangent of the oppositeangle, and the hypotenuse its secant. THEOR. II. The sides of a plane triangle are to one another asthe fines of the opposite angles. Fig. 4. From B any angle of the triangle ABC (fig. 4.), drawBD perpendicular to AC. Then, by last theorem, AB: BD:: \text{rad.}: \sin. A, \text{also } BD: BC:: \sin. C: \text{rad.} therefore ex equo inversely (GEOMETRY, Sect. III.Theor. 7.), AB: BC:: \sin. C: \sin. A. THEOR. III. The sum of any two sides of a triangle is to theirdifference as the tangent of half the sum of theangles opposite to these sides to the tangent ofhalf their difference. Fig. 5. Let ABC, fig. 5. be a triangle; AB + BC: AB- BC:: \tan. \frac{1}{2}(\angle BCA + \angle BAC): \tan. \frac{1}{2}(\angle BCA- \angle BAC). In AB produced take BE = BC, and on B as a centrewith BC or BE as a radius, describe the semicircleECF meeting AC in D; join BD, CF, and CE, andfrom F draw FG parallel to AC, meeting CE in G. Because the angles CFE, CBE, stand on the samearch CE, and the former is at the circumference of thecircle, and the latter at the centre; therefore, theangle CFE is half the angle CBE (GEOMETRY,Sect. II. Theor. XIV.); but the angle CBE is the sumof the angles BAC, BCA (GEOMETRY, Sect. I. Theor.XXIII.); therefore the angle CFE is half the sum ofthe angles BAC, BCA. Because the angle BDC is the sum of the anglesBAC, ABD, therefore the angle ABD is the differencebetween the angles BDC, BAD; but since BD = BC,the angle BDC is equal to BCD or BCA, thereforeABD is the difference of the angles BCA, BAC; butABD, or FBD, being an angle at the centre of thecircle, is double the angle FCD at the circumference,which last is equal to the alternate angle CFG; there-fore the angle CFG is half the difference of the anglesBCA, BAC. Because CE is manifestly the tangent of the angleCFE to the radius CF and CG the tangent of the angleCFG to the same radius; therefore CE: CG:: \tan.CFE: \tan. CFG, that is, CE: CG:: \tan. \frac{1}{2}(\angle BCA +\angle BAC): \tan. \frac{1}{2}(\angle BCA - \angle BAC); but because FG is pa-rallel to AC, CE: CG:: AE: AF, that is, CE: CG:: AB + BC: AB - BC, therefore AB + BC: AB- BC:: \tan. \frac{1}{2}(\angle BCA + \angle BAC): \tan. \frac{1}{2}(\angle BCA -\angle BAC). THEOR. IV. If a perpendicular be drawn from any angle of atriangle to the opposite side or base; the sum ofthe segments of the base is to the sum of theother two sides as the difference of these sidesto the difference of the segments of the base. Let ABC be a triangle (fig. 6.), and BD a perpen-dicular drawn to the base from the opposite angle;AD + DC: AB + BC:: AB - BC: AD - DC. On, B as a centre with the radius BC, describe acircle meeting AC in E, and AB in G, and the sameline produced in F. Then AC: AF:: AG: AE;now AF = AB + BC, and AG = AB - BC, and be-cause ED = DC, AE (or AD - DE) = AD - DC,therefore AC: AB + BC:: AB - BC: AD - DC. PROBLEM. Having given the sum of any two quantities and alsotheir difference, to find each of the quantities. SOLUTION. To half the sum add half the differenceof the quantities, and it will give the greater; and fromhalf the sum subtract half the difference, and it will givethe less. For let the greater of the two quantities be expressedby the line AB, and the less by BC; bisect AC in D,and take DE equal to DB, then AE = BC, and AB- BC = AB - AE = EB, and \frac{1}{2}(AB - BC) = DB;also \frac{1}{2}(AB + BC) = AD; now AB = AD + DB andBC = AD - DB, therefore the truth of the solution isevident. In a plane triangle there are five distinct parts,which are so connected with one another, that anythree of them being given, the remaining two may befound; these are, the three sides and any two of thethree angles; as to the remaining angle, that dependsentirely upon the other two, and may be found fromthem independent of the sides. If one of the angles be a right angle, then the num-ber of parts is reduced to four, and of these, any twobeing given, the remaining two may be found. Solution of the Cases of Right-angled Plane Triangles. In right-angled triangles there are four cases whichmay be resolved by the first theorem. CASE 1. The hypotenuse AC (fig. 7.) and an angleA being given, to find the sides AB, BC about theright angle. \text{SOLUTION. } \begin{cases} \text{Rad.}: \sin. A:: AC: BC, \\ \text{Rad.}: \cos. A:: AC: AB. \end{cases} Example. In the triangle ABC, let the hypotenuseAC be 144, and the angle A 39^\circ 22'. Required thesides AB and BC. To find AB. To find BC. Logarithms. Log. Rad. 10.00000 Rad. 10.00000 Sin. A 39^\circ 22' 9.80228 Cos. A 39^\circ 22' 9.88824 AC 144 2.15836 AC 144 2.15836 BC = 91.3 1.96064 AB = 111.3 2.04660 Here Plane Tri- Here the logarithms of the second and third terms gonometry are added, and the logarithm of the first term subtracted or rejected from the sum. CASE 2. A side AB, and an acute angle A (and consequently the other angle C) being given, to find the hypotenuse AC, and remaining side BC. SOLUTION. \left\{ \begin{array}{l} \text{Cof. A}: \text{rad.}:: \text{AB}: \text{AC}, \\ \text{Rad.}: \text{tan. A}:: \text{AB}: \text{BC}. \end{array} \right. Example. In the triangle ABC are given AB 208, and the angle A 35° 16', to find AC and BC. To find AC. To find BC. Cof. A 35° 16' 9.91194 Rad. 10.00000 Rad. 10.00000 Tan. A 35° 16' 9.84952 AB 208 2.31806 AB 208 2.31806 12.31806 BC=147.1 2.16758 AC=254.7 2.40612 CASE 3. The hypotenuse AC and a side AB being given, to find the angle A (and consequently C) and the side BC. SOLUTION. \left\{ \begin{array}{l} \text{AC}: \text{AB}:: \text{rad.}: \text{cos. A}, \\ \text{Rad.}: \text{sin. A}:: \text{AC}: \text{BC}. \end{array} \right. Example. Let the hypotenuse AC be 272, and the side AB 232. Required the angle A and the side BC. To find A To find BC. AC 272 2.43457 Rad. 10.00000 AB 232 2.36549 Sin. A 31° 28' 9.71767 Rad. 10.00000 AC 272 2.43457 12.36549 BC 142 2.15224 Cof. A=31° 28' 9.93092 CASE 4. The sides AB and BC about the right angle being given, to find the angle A (and thence C) and the hypotenuse AC. SOLUTION. \left\{ \begin{array}{l} \text{AB}: \text{BC}: \text{rad.}: \text{tan. A}, \\ \text{Cof. A}: \text{rad.}: \text{AB}: \text{AC}. \end{array} \right. Example. Let the side AB be 186, the side BC 152. Required the angle A, and the hypotenuse AC. To find A. To find AC. AB 186 2.26951 Cof. A 39° 15' 9.88896 BC 152 2.18184 Rad. 10.00000 Rad. 10.00000 AB 186 2.26951 12.18184 12.26951 Tan. A=39° 15' 9.91233 AC=240.2 2.38055 Solution of the Cases of Oblique-angled Triangles. In oblique-angled triangles there are also four cases, which, with their solutions, are as follows. CASE 1. Two angles A and B, and a side AB being given, to find the other sides AC, BC. VOL. XX. Part II. SOLUTION. First subtract the sum of the angles A and B from 180°, and the remainder is the angle C; then AC and BC are to be found from these proportions. \text{Sin. C}: \text{Sin. B}:: \text{AB}: \text{AC}, \text{Sin. C}: \text{Sin. A}:: \text{AB}: \text{BC}. The truth of this solution is obvious from Theor. II. Example. In the triangle ABC are given the side AB=266, the angle A 38° 40', the angle B 72° 16'; to find the sides AC and BC. First, A+B=110° 56', and 180°-110° 56'=69° 4'=C. Sin. C 69° 4' 9.97035 Sin. C 69° 4' 9.97935 Sin. B 72° 16' 9.97886 Sin. A 38° 40' 9.79573 AB 266 2.42488 AB 266 2.42488 12.40374 12.22061 AC=271.3 2.43339 BC=177.9 2.25026 CASE 2. Two sides AC, CB (fig. 9.), and the angle Fig. 9. A opposite to one of them, being given; to find the other angles B, C, and also the other side AB. SOLUTION. The angle B is found by this proportion. \text{CB}: \text{AC}:: \text{sin. A}: \text{sin. B}. When CB is less than CA, the angle B admits of two values, one of which is the supplement of the other; because, corresponding to the same value of the side AC, and the angle A, the side BC may evidently have two distinct positions, viz. CB, Cb. The angle CBA and its supplement CbA being found, the angle ACB, also the angle ACb may be found, by subtracting the sum of the two known angles from 180°, and then AB and Ab may be found by these proportions. \text{Sin. A}: \text{Sin. ACB}:: \text{CB}: \text{AB}, \text{Sin. A}: \text{Sin. ACb}:: \text{CB or Cb}: \text{Ab}. This is called the ambiguous case, on account of the angle B and the side AB having sometimes two values. This solution, like the last, is deduced from Theorem II. Example. Suppose AC 225, BC 180, and the angle A 42° 20'; to find the remaining parts. CB 180 2.25527 AC 225 2.35218 Sin. A 42° 20' 9.82830 12.18048 Sin. ABC=57° 20' 9.92521 Or sin. AbC=122 40' In the triangle ACB we have now the side AC and the angles CAB, CBA, therefore the remaining angle ACB and side AB may be found by Case 1.; and the same is true of the triangle ACb. CASE 3. Two sides CA, CB and the included angle C being given, to find the remaining angles B, A, and side AB. SOLUTION. Find AC + CB, the sum of the sides, and AC - CB their difference; also find the sum of the angles A and B (that sum is the supplement of C), and half that sum; then half the difference of the angles will be got from this proportion. (See Theor. III.). AC + CB: AC - CB:: \tan. \frac{1}{2}(B + A): \tan. \frac{1}{2}(B - A). Having now the sum and difference of the angles B and A, the angles will be found by the rule given in the problem following Theor. IV. The remaining side may be found by either of these proportions. \sin. B: \sin. C:: AC: AB; \text{ or } \sin. A: \sin. C:: BC: AB. Example. Let AC be 128, CB 90, and the angle C 48^\circ 12'. Required the remaining parts of the triangle. \begin{array}{rcl} AC + CB & 218 & 2.33846 \\ AC - CB & 38 & 1.57978 \\ \tan. \frac{1}{2}(B + A) & 65^\circ 54' & 10.34938 \\ \hline & & 11.92916 \\ \tan. \frac{1}{2}(B - A) & 21^\circ 17' & 9.59070 \end{array} Hence by the given rule in the above-mentioned problem, B = 87^\circ 11', A = 43^\circ 37'. As we now know all the angles and two sides, the remaining side may be found by Case 1. CASE 4. The three sides AB, BC and AC (fig. 10.) being given, to find the three angles A, B, C. SOLUTION. Let fall a perpendicular CD upon the greatest of the three sides from the opposite angle. Then find the difference between AD and DB by this proportion. AB: AC + CB:: AC - CB: AD - DB. The segments AD, DB may now be found severally by the rule given for finding each of the quantities whose sum and difference is given, and then the angles A and B may be found by the following proportions. \begin{aligned} CA: AD &:: \text{rad.}: \text{cos. } A, \\ CB: BD &:: \text{rad.}: \text{cos. } B. \end{aligned} The angles A, B being found, C of course is known. The first part of this solution follows from Theor. IV. The latter part from Theor. I. Example. Let AB be 125, AC 105, and BC 95. Required the angles. In this case AC + BC = 200, AC - BC = 10, therefore we have 125: 200:: 10: AD - DB = \frac{200 \times 10}{125} = 16. Now AD + DB = 125, therefore AD = 70.5 DB = 54.5. To find A. To find B. Spherical Trigonometry. AC 105 2.02119 BC 95 1.97772 AD 70.5 1.84819 BD 54.5 1.73640 Rad. 10.00000 Rad. 10.00000 11.84819 11.73640 Cos. A 47^\circ 49' 9.82700 Cos. B 55^\circ 9.75868 For the application of plane trigonometry, see MENSURATION, Sect. I. SECTION III. SPHERICAL TRIGONOMETRY. THEOR. I. If a sphere be cut by a plane through the centre, the section is a circle. THE truth of this proposition is evident from the definition of a sphere. See GEOMETRY, Sect. IX. Def. 3. DEFINITIONS. I. Any circle which is a section of a sphere by a plane passing through its centre, is called a great circle of the sphere. COR. All great circles of a sphere are equal, and the centre of the sphere is their common centre, and any two of them bisect one another. II. The pole of a great circle of the sphere is a point in the superficies of the sphere from which all straight lines drawn to the circumference of the circle are equal. III. A spherical angle is that which on the superficies of a sphere is contained by two arcs of great circles, and is the same with the inclination of the planes of these great circles. IV. A spherical triangle is a figure upon the superficies of a sphere comprehended by three arcs of great circles, each of which is less than a semicircle. THEOR. II. The arch of a great circle between the pole and the circumference of another circle is a quadrant. Let ABC be a great circle, (fig. 11.) and D its pole; let the great circle ADC pass through D, and let AEC be the common section of the planes of the two circles, which will pass through E the centre of the circle; join DA, DC. Because the chord DA is equal to the chord DC, (Def. 2.) the arc DA is equal to the arc DC; now ADC is a semicircle, therefore the arcs AD and DC are quadrants. COR. 1. If DE be drawn, the angle AED is a right angle, and DE being therefore at right angles to every line it meets with in the plane of the circle ABC, is at right angles to that plane. Therefore the straight line drawn from the pole of any great circle to the centre of the sphere is at right angles to the plane of that circle. COR. 2. The circle has two poles D, D', one on each Spherical each side of its plane, which are the extremities of a diameter of the sphere perpendicular to the plane ABC. THEOR. III. A spherical angle is measured by the arch of a great circle intercepted between the great circles containing the angle, and having the angular point for its pole. Fig. 12. Let AB, AC be two arches of great circles containing the spherical angle BAC; let BC be an arch of a great circle intercepted between them, and having A for its pole, and let BD, CD, AD be drawn to D the centre of the sphere. The arches AB, AC are quadrants, (Theor. II.); and therefore the angles ADB, ADC right angles; therefore (GEOMETRY, Sect. VII. Def. 4.), the angle BDC (which is measured by the arch BC) is the inclination of the planes of the circles BDA, CDA, and is equal to the spherical angle BAC (Def. 3.). COR. If AB, AC two arches of great circles meet in A, then A shall be the pole of a great circle passing through B and C. THEOR. IV. Two great circles whose planes are perpendicular pass through each others poles. Fig. 13. Let ACBD, AEBF be two great circles, the planes of which are at right angles to one another; from G the centre of the sphere, draw GC in the plane ACBD perpendicular to AB, then GC is also perpendicular to the plane AEBF, (GEOMETRY, Sect. VII. Theor. 12.); therefore C is the pole of the circle AEBF, and if CG be produced to D, D is the other pole of the circle AEBF. In the same manner, by drawing GE in the plane AEBF perpendicular to AB, and producing it to F, it is shewn that E and F are the poles of the circle ACBD. COR. 1. If two great circles pass through each others poles, their planes are perpendicular to one another. COR. 2. If of two great circles the first passes through the poles of the second, the second also passes through the poles of the first. THEOR. V. If the angular points of any spherical triangle be made the poles of three great circles, another triangle will be formed by their intersections, such, that the sides of the one triangle will be respectively the supplements of the measures of the angles opposite to them in the other. Fig. 14. Let the angular points of the triangle ABC be the poles of three great circles; which by their intersections form the three lunar surfaces DQ, FR, and EO; A being the pole of EF, B the pole of DF, and C the pole of ED. Then the triangle DEF which is common to three lunar surfaces will be in every respect supplemental to the triangle ABC. Spherical For let each side of ABC be produced to meet the sides that contain the angle opposite to it, in the triangle DEF; then, because BC passes through the poles of ED, DF, ED, DF must also pass through the poles of BC. (Theor. II. Cor. 2.). Therefore the points D, Q are the poles of BC. In like manner R, F are the poles of AB, and E, O the poles of AC. Hence EL, FK are quadrants, (Theor. II.); and therefore EF is the supplement of KL, but since A is the pole of EF, KL is the measure of the angle at A; thus EF is the supplement of the measure of the angle at A. In like manner FD is the supplement of the measure of the angle at B, and DE the supplement of the measure of the angle at C. Further, it will appear in the same manner that BC is the supplement of HM, the measure of the angle at D; that AB is the supplement of NK the measure of the angle at F; and that AC is the supplement of GL, the measure of the angle at E. THEOR. VI. Fig. 15. If from any point E, which is not the pole of the great circle ABC, there be drawn arches of great circles EA, EK, EB, &c. the greatest of these is EGA, which passes through G the pole of ABC, and EC the remainder of the semicircle is the least, and of the other, EK, EB, &c. EK which is nearer to EA is greater than EB, which is more remote. Let AC be the common section of the planes of the great circles AEC, ABC; draw EH perpendicular to AC, which will be perpendicular to the plane of the circle ABC (GEOMETRY, Sect. VII. Theor. XII.) and join AE, KE, BE, KH, BH. Then of all the straight lines drawn from H to the circumference, HA is the greatest, HC the least, and HK greater than HB. Therefore in the right-angled triangles EHA, EHK, EHB, EHC, which have the side EH common, EA is the greatest hypotenuse, EC the least, and EK greater than EB, consequently the arch EGA is the greatest, EC the least and EK greater than EB. THEOR. VII. Any two sides of a spherical triangle are together greater than the third, and all the three sides are together less than a circle. Let ABC be a spherical triangle, let D be the centre of the sphere, join DA, DB, DC. The solid angle at D is contained by three plane angles ADB, BDC, ADC, any two of which are greater than the third, (GEOMETRY, Sect. VII. Theor. XV.); and therefore any two of the arches AB, BC, AC which measure these angles must be greater than the third arch. To prove the second part of the proposition, produce the sides AB, AC until they meet again in E; then ECA and EBA are semicircles; now CB is less than CE + EB, therefore CB + CA + BA is less than CE + EB + CA + BA, but these four arches make up two semicircles; therefore CB + CA + BA is less than a circle. THEOR. VIII. If two sides of a spherical triangle be equal, the angles opposite to them are equal, and conversely. In the triangle ABC, if the sides AB, AC be equal, the angles ABC, ACB are also equal. If AB, AC be quadrants, ABC, ACB are right angles. If not, let the tangent to the side AB at B meet EA the line of common section of the planes AB, AC in F, and let the tangents to the base BC at its extremities meet each other in G; also, let FC, FG, EC, and EB be joined. Then the triangles FEB, FEC have FE common, EB=EC, and the angle AEB=AEC, therefore FB=FC, and the angle FCE=FBE a right angle: hence FC is a tangent, and the triangles FGB, GCF are mutually equilateral, therefore the angle FBG=FCG and consequently the spherical angle ABC=ACB. Again, if the angles ABC, ACB be equal, the side AB=AC. For if in fig. 14. the angle ABC be equal to ACB, the side DF of the supplemental triangle DEF will be equal to the side DE (Theor. V.); therefore the angle DEF=DFE, and consequently in the triangle ABC, the side AC=AB by Theorem V. COR. In any triangle the greater angle is subtended by the greater side; and conversely. For if the angle ACB be greater than ABC (fig. 18.) let BCD=ABC, then BD=DC, and AB=AD+DC, which is greater than AC (Theor. VII.). The converse is demonstrated in the same manner as the like property of plane triangles, (GEOMETRY, Sect. I. Theor. XIII.). THEOR. IX. All the angles of a spherical triangle are together greater than two, and less than six right angles. In the triangle ABC (fig. 14.) the three angles are together less than six right angles, because when added to the three exterior angles they only make six; and they are greater than two right angles, because their measures GH, KL, MN, added to DE, EF, FD, are equal to three semicircles; and DE, EF, FD being less than two semicircles (Theor. VII.) GH, KL, MN must be greater than one. THEOR. X. Any two angles of a spherical triangle are together greater, equal, or less than two right angles, according as the sum of the opposite sides is greater, equal, or less than a semicircle; and conversely. Let the sides AB, AC (fig. 19.) of the spherical triangle ABC be produced to meet in D; then it is evident that according as the sum of AB, BC is greater, equal, or less than the semicircle ABD, the side BC will be greater, equal, or less than BD; the angle D or A will be greater, equal, or less than BCD, and the sum of the angles BAC, BCA greater, equal, or less than the sum of BCA, BCD, which is two right angles. COR. According as half the sum of any two sides of a spherical triangle is greater, equal, or less than a quadrant, half the sum of the opposite angles will be greater, equal, or less than a right angle. THEOR. XI. In a right-angled triangle, according as either of the sides about the right angle is greater, equal, or less than a quadrant, its opposite angle is greater, equal, or less than a right angle; and conversely. Let ABC (fig. 20.) be a triangle right-angled at B, and let the sides AB, BC be produced to meet in D; then, because they pass through each other's poles, E the middle point of BAD will be the pole of BCD; let a great circle pass through the points CE. The arch EC is a quadrant, and the angle ECB a right angle. Now it is plain, that according as AB is greater, equal, or less than the quadrant EB, the opposite angle ACB will be greater, equal, or less than the right angle ECB, and conversely. COR. 1. If the two sides be both greater, or both less than quadrants, the hypotenuse will be less than a quadrant; but if the one be greater and the other less, the hypotenuse will be greater than a quadrant, and conversely. For in the triangles ABC, ADC, right-angled at B, D, in which the sides AB, BC are less, and consequently AD, DC greater than quadrants, the hypotenuse AC is less than a quadrant, because it is nearer to CB than the quadrant CE. But in the triangle ABC, of which the side AB is greater, and BC less than a quadrant, the hypotenuse AC is greater than a quadrant, because it is further from CB than CE is. COR. 2. In every spherical triangle, of which the two sides are not both quadrants, if the perpendicular from the vertex fall within, the angles at the base will be both acute or both obtuse; but if it fall without, the one will be obtuse, and the other acute; and conversely. THEOR. XII. In any right-angled spherical triangle, as radius is to the sine of the hypotenuse, so is the sine of one of the oblique angles to the sine of its opposite side. Let ABC (fig. 21.) be a spherical triangle, having a right angle at B; and let AD, BD, CD be drawn to the centre of the sphere. From C, in the plane DCA, let CE be drawn perpendicular to DA, and from E, in the plane DBA, draw EF perpendicular to the same line, and let CF be joined. Then because DA is perpendicular to the two lines CE, EF, it is perpendicular to the plane CEF, and consequently the plane CEF is perpendicular to the plane DBA; but the plane DCB is also perpendicular to DBA; therefore their line of common section CF is perpendicular to the same: Hence CFD, CFE are right angles. Now in the right-angled triangle CFE, \text{rad.}: CE:: \text{fin.} E: CF; but the angle CEF, being the inclination of the planes DCA, DBA, is the same with the spherical angle CAB, CE is the sine of AC, and CF the sine of BC; therefore \text{rad.}: \text{fin.} AC:: \text{fin.} A: \text{fin.} BC. COR. Spherical Trigonometry. COR. 1. As radius to the cosine of either of the sides, so is the cosine of the other to the cosine of the hypothenuse. Fig. 22. For let the great circle of which A is the pole, meet the three sides in D, E, F; then F is the pole of AD; and applying this proposition to the complementary triangle FCE, rad.: sin. FC:: sin. F: sin. CE; that is, rad.: cos. BC:: cos. AB: cos. AC. COR. 2. As radius to the cosine of one of the sides, so is the sine of its adjacent angle to the cosine of the other angle. THEOR. XIII. In any right-angled triangle, as radius to the sine of one of the sides, so is the tangent of the adjacent angle to the tangent of the other side. Fig. 23. From B let BE be drawn perpendicular to DA, and from E, EF also perpendicular to DA, in the plane DCA, to meet DC in F, and let BF be joined. It may be shown as in the preceding proposition, that FB is perpendicular to the plane DBA; hence FB is the tangent of BC, and FBE is a right-angled triangle; therefore rad.: EB:: tan. E: FB; that is rad.: sin. AB:: tan A: tan. BC. COR. 1. As radius to the cosine of the hypothenuse, so is the tangent of one of the angles to the cotangent of the other. For, in the complementary triangle FCE, (fig. 22.) rad.: sin. CE:: tan. C: tan. FE, that is, rad.: cos. AC:: tan. C: cot. A, or, rad.: cos. AC:: tan. A: cot. C. COR. 2. As radius is to the cosine of one of the angles, so is the tangent of the hypothenuse to the tangent of the side adjacent to that angle. For rad.: sin. FE:: tan F: tan. CE; that is, rad.: cos. A:: cot. AB: cot. AC, or rad.: cos. A:: tan. AC: tan. AB. Napier's Rule for Circular Parts. Let the hypothenuse, the two angles, and the complements of the two sides of any right-angled spherical triangle be called the five circular parts of the triangle. Any one of these being considered as the middle part, let the two which are next to it be called the adjacent parts, and the remaining two the opposite parts. Then the two preceding theorems, with their corollaries, may be all expressed in one proposition adapted to practice, as follows. In any right-angled spherical triangle, the rectangle under radius, and the cosine of the middle part, is equal to the rectangle under the cotangents of the adjacent parts, or to the rectangle under the sines of the opposite parts. Fig. 24. CASE 1. Let the hypothenuse AC be the middle part. Then, rad.: cos. AC:: tan. C: cot. A (Theor. 13. Cor. 1.). Therefore (rad.: tan. C::) cot. C: rad.:: cos. AC: cot. A. And rad.: cos. AB:: cos. BC: cos. AC (Theor. 12. Cor. 1.). CASE 2. Let the angle A be the middle part. Then (Theor. 13. Cor. 2.) rad.: cos. A:: tan. AC: tan. AB. Therefore, (rad.: tan. AC::) cot. AC: rad.:: cos. A: tan. AB. And (Theor. 12. Cor. 2.) rad.: cos. BC:: sin. C: cos. A. CASE 3. Let the complement of the side AB be the middle part. Then (Theor. 13.) rad.: sin. AB:: tan. A: tan. BC. Therefore (rad.: tan. A::) cot. A: rad.:: sin. AB: tan. BC. And (Theor. 12.) rad.: sin. AC:: sin. C: sin. AB. We are indebted for the foregoing rule to Napier, the celebrated inventor of logarithms. It comprehends all the propositions which are necessary for the resolution of right-angled triangles, and being easily remembered, is perhaps one of the happiest instances of artificial memory that is known. THEOR. XIV. In any spherical triangle, the sines of the sides are proportional to the sines of the opposite angle. This proposition has been demonstrated in the case of right-angled triangles. Let ABC be any oblique-angled triangle, divided into two right-angled triangles, ABD, CBD, by the perpendicular BD, falling from the vertex upon the base AC. In the former, the complement of BD being the middle part, rad. \times sin. BD = sin. AB \times sin. A, (NAPIER'S RULE). In the latter, the complement of BD being the middle part, rad. \times sin. BD = sin. BC \times sin. C. Hence sin. AB \times sin. A = sin. BC \times sin. C, and sin. AB: sin. BC:: sin. A: sin. C. COR. 1. The cosines of the two sides are to one another directly as the cosines of the segments of the base. This is proved by making AB, BC the middle part. COR. 2. The tangents of the two sides are to one another inversely as the cosines of the vertical angles. This will follow from making the angles ABD, CBD the middle parts. LEMMA 1. The sum of the tangents of two arcs is to their difference, as the rectangle under the sine and cosine of half their sum to the rectangle under the sine and cosine of half their difference. For, putting a and b for any two arcs, by the arithmetic of sines (ALGEBRA, § 353.), \sin. a \cos. b + \cos. a \sin. b = \sin. (a+b). Let each side of this equation be divided by \cos. a \cos. b, and we get \frac{\sin. a}{\cos. a} + \frac{\sin. b}{\cos. b} = \frac{\sin. (a+b)}{\sin. a \cos. b}. that is, \tan. a + \tan. b = \frac{\sin. (a+b)}{\sin. a \cos. b}. In like manner, from the formula \sin. (a-b) = \sin. a \cos. b - \cos. a \sin. b, we get \tan. a - \tan. b = \frac{\sin. (a-b)}{\sin. a \cos. b}. therefore \tan. a + \tan. b: \tan. a - \tan. b:: \sin. (a+b): \sin. (a-b), and remarking that \sin. (a+b) = 2 \sin. \frac{1}{2}(a+b) Spherical \frac{1}{2}(a+b) \cos. \frac{1}{2}(a+b), and \sin. (a-b) = 2 \sin. \frac{1}{2}(a-b) \cos. \frac{1}{2}(a-b), (ALGEBRA, § 358) it follows that \tan. a + \tan. b: \tan. a - \tan. b:: \sin. \frac{1}{2}(a+b) \cos. \frac{1}{2}(a+b): \sin. \frac{1}{2}(a-b) \cos. \frac{1}{2}(a-b). LEMMA 2. The sum of the sines of two arcs is to their difference, as the rectangle under the sine of half the sum and cosine of half the difference of these arcs is to the rectangle under the sine of half the difference and cosine of half the sum. For it has been shown in the arithmetic of sines (ALGEBRA, § 355), that \sin. (p+q) + \sin. (p-q) = 2 \sin. p \cos. q, \sin. (p+q) - \sin. (p-q) = 2 \cos. p \sin. q. Let p = \frac{1}{2}a + \frac{1}{2}b, and q = \frac{1}{2}a - \frac{1}{2}b, so that p+q = a and p-q = b, then these formulas become \sin. a + \sin. b = 2 \sin. \frac{1}{2}(a+b) \cos. \frac{1}{2}(a-b) \sin. a - \sin. b = 2 \cos. \frac{1}{2}(a+b) \sin. \frac{1}{2}(a-b). Therefore, \sin. a + \sin. b: \sin. a - \sin. b:: \sin. \frac{1}{2}(a+b) \cos. \frac{1}{2}(a-b): \cos. \frac{1}{2}(a+b) \sin. \frac{1}{2}(a-b). LEMMA 3. The sum of the sines of two arcs is to their difference, as the tangent of half the sum of these arcs is to the tangent of half their difference. For, dividing the latter antecedent and consequent of the proposition in the foregoing lemma by \cos. \frac{1}{2}(a+b) \times \cos. \frac{1}{2}(a-b), we have \sin. a + \sin. b: \sin. a - \sin. b:: \frac{\sin. \frac{1}{2}(a+b)}{\cos. \frac{1}{2}(a+b)} \frac{\sin. \frac{1}{2}(a-b)}{\cos. \frac{1}{2}(a-b)}, that is, because \frac{\sin. \frac{1}{2}(a+b)}{\cos. \frac{1}{2}(a+b)} = \tan. \frac{1}{2}(a+b) and \frac{\sin. \frac{1}{2}(a-b)}{\cos. \frac{1}{2}(a-b)} = \tan. \frac{1}{2}(a-b). LEMMA 4. The sum of the cosines of two arcs is to their difference, as the cotangent of half the sum of these arcs is to the tangent of half their difference. By Arithmetic of sines (ALGEBRA, § 355), \cos. (p-q) + \cos. (p+q) = 2 \cos. p \cos. q, \cos. (p-q) - \cos. (p+q) = 2 \sin. p \sin. q. Let p = \frac{1}{2}(b+a) and q = \frac{1}{2}(b-a), then p-q = a and p+q = b, and the two formulas become \cos. a + \cos. b = 2 \cos. \frac{1}{2}(b+a) \cos. \frac{1}{2}(b-a), \cos. a - \cos. b = 2 \sin. \frac{1}{2}(b+a) \sin. \frac{1}{2}(b-a); Hence, \cos. a + \cos. b: \cos. a - \cos. b:: \cos. \frac{1}{2}(b+a) \cos. \frac{1}{2}(b-a): \sin. \frac{1}{2}(b+a) \sin. \frac{1}{2}(b-a); and dividing the latter antecedent and consequent by \sin. \frac{1}{2}(b+a) \cos. \frac{1}{2}(b-a), \cos. a + \cos. b: \cos. a - \cos. b:: \frac{\cos. \frac{1}{2}(b+a)}{\sin. \frac{1}{2}(b+a)}: \frac{\sin. \frac{1}{2}(b-a)}{\cos. \frac{1}{2}(b-a)}, \text{ that is, because } \frac{\cos. \frac{1}{2}(b+a)}{\sin. \frac{1}{2}(b+a)} = \cot. \frac{1}{2}(b+a) and \frac{\sin. \frac{1}{2}(b-a)}{\cos. \frac{1}{2}(b-a)} = \tan. \frac{1}{2}(b-a), we have \cos. a + \cos. b: \cos. a - \cos. b:: \cot. \frac{1}{2}(b+a): \tan. \frac{1}{2}(b-a). In the demonstration of the remaining theorems, we shall put A, B for the angles A and B at the base of the spherical triangle ACB (fig. 26), a and b for the sides opposite to these angles, p and q for the segments of the base BD, AD made by the perpendicular arch CD, P and Q for the vertical angles BCD, ACD; we shall also put s for \frac{1}{2}(a+b), d for \frac{1}{2}(a-b), s' for \frac{1}{2}(p+q), d' for \frac{1}{2}(p-q), S for \frac{1}{2}(A+B), D for \frac{1}{2}(A-B), S' for \frac{1}{2}(P+Q), and D' for \frac{1}{2}(P-Q). THEOR. XV. In any spherical triangle, the tangent of half the sum of the segments of the base is to the tangent of half the sum of the two sides, as the tangent of half their difference to the tangent of half the difference of the segments of the base. For by Theor. XIV. Cor. 1. \cos. a: \cos. b:: \cos. p: \cos. q; therefore, \cos. a + \cos. b: \cos. a - \cos. b:: \cos. p + \cos. q: \cos. p - \cos. q, hence (Lemma 4.) \cos. s: \tan. d:: \cos. s': \tan. d', or \cos. s: \cos. s':: \tan. d: \tan. d'; but \cos. s: \cos. s':: \tan. s': \tan. s, therefore, \tan. s': \tan. s:: \tan. d: \tan. d'. This proposition expressed in words at length is the theorem to be demonstrated. THEOR. XVI. The cotangent of half the sum of the vertical angles and the tangent of half their difference, or the cotangent of half their difference and the tangent of half their sum, according as the perpendiculars fall within or without, are reciprocally proportional to the tangents of half the sum and half the difference of the angles at the base. For, taking the case in which the perpendicular CD (fig. 27.) falls within, let EFG be the supplemental triangle, let the arcs GE, GF meet again in I, and produce CA, CB to meet EF in H and K. Because G and I are the poles of AB, the perpendicular CD, if produced, will pass through G and I; let it meet EF in I; then, because C is the pole of EF, the arch GCI is perpendicular to EF, and since E is the pole of BC, KE = a quadrant = FH, and EH = KF, and IF = IE = IK = IH. In the triangle LEF, by the preceding proposition, \tan. \frac{1}{2}(FI+IE): \tan. \frac{1}{2}(FI+LE):: \tan. \frac{1}{2}(FI+LE): \tan. \frac{1}{2}(FI+IE) or \tan. \frac{1}{2}(KI+IH): \tan. \frac{1}{2}(FI+LE), or FE, being the supplement of C, (Theor. 5.), \tan. \frac{1}{2}FE = \cot. \frac{1}{2}C; and FI, LE being the supplements of FG and GE, FL and LE are the measures of the angles A, B; moreover, IK, IH are the measures of the angles BCD, ACD, therefore, \cot. \frac{1}{2}C, or \cot. \frac{1}{2}(P+Q): \tan. \frac{1}{2}(A+B): \tan. \frac{1}{2}(A-B): \tan. \frac{1}{2}(P-Q). In the very same way it may be proved, when the perpendicular falls without the triangle, that \cot. \frac{1}{2}(P-Q): \tan. \frac{1}{2}(A+B):: \tan. \frac{1}{2}(A-B): \tan. \frac{1}{2}(P+Q). THEOR. XVII. In any spherical triangle, the sine of half the sum of the sides is to the sine of half their difference, as the cotangent of half the vertical angle to the tangent of half the difference of the angles at the base. For since \tan. a: \tan. b:: \cos. Q: \cos. P, therefore, \tan. \frac{1}{2}(a+b): \tan. \frac{1}{2}(a-b):: \cos. Q: \cos. P. Spherical trigonometry. \tan. a + \tan. b: \tan. a - \tan. b:: \cot. Q + \cot. P: \cot. Q - \cot. P; hence, by Lemma 2 and 4. \sin. s \cos. s: \sin. d \cos. d:: \cot. S': \tan. D' \dots (1). Again, because (by Theor. XIV.) \sin. a: \sin. b:: \sin. A: \sin. B, therefore, \sin. a + \sin. b: \sin. a - \sin. b:: \sin. A + \sin. B: \sin. A - \sin. B; hence, (by Lemma 2, and 3.) \sin. s \cos. d: \sin. d \cos. s:: \tan. S: \tan. D \dots (2). Taking now the product of the corresponding terms of the proportions (1) and (2), and rejecting the factor \cos. s \cos. d, which is common to the first antecedent and consequent of the resulting proportion, we have, \sin. s: \sin. d:: \cot. S' \tan. S: \tan. D' \tan. D. But since by Theor. XVI. \tan. S: \tan. D':: \cot. S': \tan. D, therefore \cot. S' \tan. S: \tan. D' \tan. D:: \cot. S': \tan. D; therefore, \sin. s: \sin. d:: \cot. S': \tan. D, and \sin. s: \sin. d:: \cot. S': \tan. D, this proportion when expressed in words is the proportion to be demonstrated. THEOR. XVIII. In any spherical triangle, the cosine of half the sum of the two sides is to the cosine of half their difference, as the cotangent of half the vertical angle to the tangent of half the sum of the angles at the base. For it has been proved in last theorem that \sin. s \cos. s: \sin. d \cos. d:: \cot. S': \tan. D' \\ \sin. s \cos. d: \sin. d \cos. s:: \tan. S: \tan. D; therefore, dividing the terms of the first of these two proportions by the corresponding terms of the second, we get, \frac{\cos. s}{\cos. d}: \frac{\cos. d}{\cos. s}:: \frac{\cot. S'}{\tan. S}: \frac{\tan. D'}{\tan. D}. Hence, multiplying the first and second terms by \cos. s \times \cos. d, and the third and fourth by \tan. S \tan. D, we have, \cos. s: \cos. d:: \cot. S' \tan. D: \tan. S \tan. D'. But since by Theor. XVI. \tan. D: \tan. D':: \cot. S': \tan. S, therefore, \cot. S' \tan. D: \tan. S \tan. D':: \cot. S': \tan. S; therefore, \cos. s: \cos. d:: \cot. S': \tan. S, and \cos. s: \cos. d:: \cot. S': \tan. S. THEOR. XIX. In any spherical triangle, the sine of half the sum of the angles at the base is to the sine of half their difference, as the tangent of half the base to the tangent of half the difference of the two sides. For the same construction being made as in Theor. XVI. in the triangle ELF (fig. 27.) \sin. \frac{1}{2}(FL+LE): \sin. \frac{1}{2}(FL-LE):: \cot. \frac{1}{2}L: \tan. \frac{1}{2}(E-F) (Theor. XVII.); but EFG being the supplemental triangle of ABC, LF and LE are the measures of A and B, L is the supplement of AB, and LFE, LEF are the measures of the sides AC, BC (Theor. V.); therefore \sin. \frac{1}{2}(A+B): \sin. \frac{1}{2}(A-B):: \tan. \frac{1}{2}AB: \tan. \frac{1}{2}(BC-AC). THEOR. XX. In any spherical triangle, the cosine of half the sum of the angles at the base is to the cosine of half their difference, as the tangent of half the base to the tangent of half the sum of the two sides. For in the triangle ELF, \cos. \frac{1}{2}(LF+LE): \cos. \frac{1}{2}(LF-LE):: \cot. \frac{1}{2}L: \tan. \frac{1}{2}(E+F) (Th. XVIII.) that is, because of the relation of the triangle FLE to ABC, as expressed in last theorem, \cos. \frac{1}{2}(A+B): \cos. \frac{1}{2}(A-B):: \tan. \frac{1}{2}AB: \tan. \frac{1}{2}(BC+AC). SCHOLIUM. Let one of the six parts of any spherical triangle be neglected; let the one opposite to it, or its supplement, if an angle, be called the middle part, the two next to it the adjacent parts, and the remaining two the opposite parts. Then the four preceding propositions, which are called Napier's Analogies, because first invented by him, may be included in one, as follows. In any spherical triangle, the sine or cosine of half the sum of the adjacent parts, is to the sine or cosine of half their difference, as the tangent of half the middle part to the tangent of half the difference or half the sum of the opposite parts, that is, \sin. \frac{1}{2}(A+a): \sin. \frac{1}{2}(A-a):: \tan. \frac{1}{2}M: \tan. \frac{1}{2}(O-o). \cos. \frac{1}{2}(A+a): \cos. \frac{1}{2}(A-a):: \tan. \frac{1}{2}M: \tan. \frac{1}{2}(O+o). When A, a and M are given, by the first proportion, \frac{1}{2}(O-o) is found, and by the second \frac{1}{2}(O+o); thence O and o may be had immediately by the problem following Theor. IV. PLANE TRIGONOMETRY. The Cases of Right-angled Spherical Triangles. In a right-angled triangle, let c denote the side opposite the right angle, a, b the sides containing it, and A, B the opposite angles, A being opposite to a, and B to b. Then, combining these quantities two by two, there will be found to be six distinct combinations, or cases. CASE 1. When c, A, the hypotenuse and one of the angles are given; to find a, b, B.a is found by Theor. XII.; b by Theor. XIII. Cor. 2. and B by Theor. XIII. Cor. 1. CASE 2. Given a, B, a side and its adjacent angle. Sought, A, b, c.A is found by Theor. XII. Cor. 2.; b by Theor. XIII.; c by Theor. XIII. Cor. 2. CASE 3. Given a, A, a side and its opposite angle; to find b, B, c.b is found by Theor. XIII.; B by Theor. XII.; Cor. 2. c by Theor. XII. CASE 4. Given c, a, the hypotenuse, and one of the sides; to find A, b, B.A is found by Theor. XII.; b by Theor. XII. Cor. 1.; B by Theor. XIII. Cor. 2. CASE. CASE 5. Given a, b, the two sides. Sought A, B, c.A is found by Theor. XIII.; B by the same; c byTheor. XII. Cor. 1. CASE 6. Given A, B, the two angles. Soughta, b, c.a and b are found by Theor. XII. Cor. 3; 2c byTheor. XIII. Cor. 1. THE cases may be all resolved also by Napier's Rules,observing to make each of the things given the middlepart: then two of the required parts will be found, andthe remaining part is found by making it the middlepart. By Theor. II. and Cor. 1. each of the unknown partsis, in every case except the third, limited to onevalue. The Cases of Oblique-angled Spherical Triangles. In any spherical triangle let the sides be denoted bya, b, c, and the opposite angles by A, B, C respec-tively. Let p, q denote the segments into which a side is di-vided by a perpendicular from the opposite angle, andP, Q the parts into which it divides the angle. Com-bining the six quantities a, b, c, A, B, C, three bythree, there are found six distinct combinations orcases. CASE 1. Given a, A, b, two sides and an angle op-posite to one of them. Sought c, B, C. B is found by Theor. XIV.; c by either Theor. XIX.;or Theor. XX.; C by Theor. XVII. or Theor. XVIII. CASE 2. Given A, a, B, two angles and a side op-posite to one of them. Sought b, c, C.b is found by Theor. XIV.; c and C as in Case 1. CASE 3. Given a, C, b, two sides and the includedangle. Sought A, B, c. Find \frac{1}{2}(A-B) by Theor. XVII. and \frac{1}{2}(A+B) by SphericalTheor. XVIII. and thence A and B by the rule Trigonome-try. SECT. II. for finding each of two quantities whose sumand difference are given. All the angles being known,also two sides, c is found by Theor. XIV. CASE 4. Given A, c, B, two angles and a side be-tween them. Sought a, B, b. Find \frac{1}{2}(a-b) by Theor. XIX. and \frac{1}{2}(a+b) byTheor. XX. and thence a, b. All the sides and twoangles being now known, C is found by Theor. XIV. CASE 5. Given a, b, c, the three sides. Sought A, B, C. Draw a perpendicular from any one of the angles,dividing the opposite side into the segments p, q. Find\frac{1}{2}(p-q) by Theor. XV. and then, from \frac{1}{2}(p+q) and\frac{1}{2}(p-q), find p, q. The triangle being now resolvedinto two right-angled triangles, the angles may be foundby Case 4. of right-angled triangles. CASE 6. Given A, B, C, the three angles. Soughta, b, c. Draw a perpendicular, dividing any one of the anglesinto the parts P, Q. Find \frac{1}{2}(P-Q) by Theor. XVI.and then P, Q. The triangle being now resolved intotwo right-angled triangles, the sides may be found byCase 6. of right-angled triangles. By Theor. X. XI. and Cor. each of the unknownparts is limited to one value in all the cases, except insome of the subcases of the first and second. As every oblique-angled triangle may be resolved in-to two right-angles, all these cases may be resolved bymeans of Napier's Rule, and the 15th proposition only.And the cases may be reduced to three, by using thesupplemental triangle. TRI TRIHLATÆ, from tres, "three," and hilum,"an external mark on the seed;" the name of the 23dclass in Linnaeus's Fragments of a Natural Method;consisting of plants with three seeds, which are markedwith an external cicatrix or scar, where they are fastenedwithin the fruit. See BOTANY. TRIM, implies in general the state or disposition bywhich a ship is best calculated for the several purposes ofnavigation. Thus the trim of the hold denotes the most conven-ient and proper arrangement of the various materials con-tained therein relatively to the ship's motion or stabilityat sea. The trim of the masts and sails is also their mostopposite situation with regard to the construction of theship and the effort of the wind upon her sails. See SEA-MANSHIP. TRINGA, SANDPiper; a genus of birds belong- TRI ing to the order of grallæ. See ORNITHOLOGYIndex.